SOLUTION: Find the equation for the tangent line to the function f(x) = 4x3+3x2+5x+6 when x = 2. Put your answer in slope intercept (y= mx+ b) format, and use it to identify the y-interc

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Question 856120: Find the equation for the tangent line to the function f(x) = 4x3+3x2+5x+6 when x = 2.
Put your answer in slope intercept (y= mx+ b) format, and use it to identify the y-intercept for this line.
The y-intercept for the tangent line to the function f(x) = 4x3+3x2+5x+6 when x = 2 is ________________________?
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
if i understand this question correctly, you are looking for the derivative of the function which tells you the slope of the tangent line and then you want to find the equation of the tangent line which will include the y-intercept if that equation is in slope intercept form.

start with:
4x^3 + 3x^2 + 5x + 6
find dy/dx which is equal to:
12x^2 + 6x + 5
when x = 2, dy/dx tells you what the slope of the tangent line is.
replace x with 2 in that equation and you get:
12x^2 + 6x + 5 equals:
12(2^2) + 6(2) + 5 which equals:
48 + 12 + 5 which equals 65.

65 is the slope of the tangent line to your original equation at x = 2.
now that you have the slope, you want to find the equation of the tangent line.

find the value of y when x = 2 in your original equation.
replace x with 2 in your original equation to get:
4x^3 + 3x^2 + 5x + 6 equals:
4(2^3) + 3(2^2) + 5(2) + 6 which equals:
32 + 12 + 10 + 6 which equals 60.

when x = 2, your original equation passes through the point (2,60).
the tangent line to your original equation also passes through this point.

the slope intercept form of the equation for a straight line is:
y = mx + b where:
m = the slope and b is the y-intercept.

you know the slope of the tangent line is 65, so this equation becomes:
y = 65x + b
you know the equation passes through the point (2,60), so you can replace y with 60 and x with 2 and find the value of b.
y = 65x + b becomes:
60 = 65(2) + b
simplify to get:
60 = 130 + b
subtract 130 from both sides of this equation to get:
b = 60 - 130 which makes b = -70

the equations of the tangent line to your original equation at x = 2 is:
y = 65x - 70.

the slope is 65.
the y-intercept is -70.

a graph of your original equation and the tangent line to that equation at x = 2 is shown below:

the horizontal line at x = 60 shows you where the intersection point between the original equation and the tangent line is.
trace a vertical line down to the x-axis and it will intersect the x-axis at x = 3.